Question

The airspeed of an airplane is 600 km/hr, and there is a wind blowing northeast at 90 km/hr. If the airplane needs to head due north, at what angle does the pilot need to fly the plane?

Answer 1

The pilot has to fly the plane at an angle 83° north of west.

Step-by-step explanation:

The airspeed of an airplane is 600 km/hr ( north direction)

In figure,
`V_(PA)=600\text{ km/h}`

A wind blowing northeast at 90 km/h

In figure,
`V_(A)=90\text{ km/h}`

Let actual speed of airplane and direction of plane,

Speed = x km/h

Direction = Ф (north of west, refer figure)

Now, we divide each velocity in horizontal and vertical component.

• For wind,
  `V_(A)=90\text{ km/h}`

Horizontal component,
`V_{\text{Horizontal}}=90\cos45^\circ\text{ km/h}`

Vertical component,
`V_{\text{Vertical}}=90\sin45^\circ\text{ km/h}`

• For airplane,

Horizontal component,
`V_{\text{Horizontal}}=x\cos\theta\text{ km/h}`

Vertical component,
`V_{\text{Vertical}}=x\sin\theta\text{ km/h}`

If the airplane needs to head due north.

Horizontal component must be cancel out.

Therefore,
`x\cos\theta=90\cos45^\circ----------(1)`

Speed of airplane is 600 km/h due to north.

So, Sum of vertical component must be equal to 600

Therefore,
`x\sin\theta+90\sin45^\circ=600-----------(2)`

Using eq(1) and eq(2) we get,

`\tan\theta=(600√(2)-90)/(90)`

`\theta=83.23^\circ`

`\theta\approx 83^\circ`

Hence, The pilot has to fly the plane at an angle 83° north of west.